1
$\begingroup$

Show that a product of unitary (orthogonal) matrices is unitary (orthogonal) as well.

This is one of my review problems, but I am having difficulty starting because my definition of a unitary operator is that it is an invertible isometry. So this tells us, for some matrix U, U* U = I. But I guess I am sort of lost as to where to go from there.

Are you able to use the fact that, for unitary matrices A and B, |Det(A)|=1, so |Det(AB)|=|Det(A)||Det(B)|=(1)(1)=1. So |Det(AB)|=1, which is a property of unitary matrices. Or is this not enough?

$\endgroup$
3
  • $\begingroup$ That is not enough because while a unitary matrix has determinant 1, there are matrixes with determinant 1 that are not unitary. $\endgroup$
    – A. Thomas Yerger
    Nov 2, 2016 at 6:01
  • $\begingroup$ So then how could one make a complete proof? Would it still involve that |Det(A)|=1 concept or is that not on the right track? $\endgroup$
    – readyforthebettys
    Nov 2, 2016 at 6:03
  • $\begingroup$ If $A$ and $B$ are unitary and you want to show $AB$ is unitary, then you should show $AB(AB)^{*} = (AB)^{*}AB = I$. $\endgroup$
    – JessicaK
    Nov 2, 2016 at 6:05

1 Answer 1

Reset to default
2
$\begingroup$

Unitary matrixes have the property that $UU^* = U^*U = I$. Given that $U$ and $T$ have this property, you want to show $UT$ also has this property. So consider $(UT)(UT)^*$ The conjugate transpose has the property that it is order-reversing under multiplication. This is the key. So $(UT)(UT)^* = UTT^*U^* = UIU^* = UU^* = I,$ which is what you wanted to prove. No determinants required.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .